Date | November 2011 | Marks available | 3 | Reference code | 11N.1.sl.TZ0.10 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 10 | Adapted from | N/A |
Question
A bag contains 7 red discs and 4 blue discs. Ju Shen chooses a disc at random from the bag and removes it. Ramón then chooses a disc from those left in the bag.
Write down the probability that
(i) Ju Shen chooses a red disc from the bag;
(ii) Ramón chooses a blue disc from the bag, given that Ju Shen has chosen a red disc;
(iii) Ju Shen chooses a red disc and Ramón chooses a blue disc from the bag.
Find the probability that Ju Shen and Ramón choose different coloured discs from the bag.
Markscheme
(i) \(\frac{7}{{11}}\) (\(0.636\), \(63.6\% \)) (\(0.636363 \ldots \)) (A1) (C1)
(ii) \(\frac{4}{{10}}\) \(\left( {\frac{2}{5}{\text{, }}0.4{\text{, }}40\% } \right)\) (A1) (C1)
(iii) \(\frac{{28}}{{110}}\) \(\left( {\frac{{14}}{{55}}{\text{, }}0.255{\text{, }}25.5\% } \right)\) \(0.254545 \ldots \) (A1)(ft) (C1)
Note: Follow through from the product of their answers to parts (a) (i) and (ii).
[3 marks]
\(\frac{{28}}{{110}} + \left( {\frac{4}{{11}} \times \frac{7}{{10}}} \right)\) OR \(2 \times \frac{{28}}{{110}}\) (M1)(M1)
Notes: Award (M1) for using their \(\frac{{28}}{{110}}\) as part of a combined probability expression. (M1) for either adding \({\frac{4}{{11}} \times \frac{7}{{10}}}\) or for multiplying by 2.
\( = \frac{{56}}{{110}}\) \(\left( {\frac{{28}}{{55}}{\text{, }}0.509{\text{, }}50.9\% } \right)\) (\(0.509090 \ldots \)) (A1)(ft) (C3)
Note: Follow through applies from their answer to part (a) (iii) and only when their answer is between 0 and 1.
[3 marks]
Examiners report
The vast majority of candidates were able to pick up the first two marks by confidently identifying the number of favourable outcomes/total number of outcomes. Difficulties arose however when combining events and only the more able candidates were able to progress successfully with the remainder of the question. As usual in this type of question, there was an abundance of incorrect answers greater than 1 given.
The vast majority of candidates were able to pick up the first two marks by confidently identifying the number of favourable outcomes/total number of outcomes. Difficulties arose however when combining events and only the more able candidates were able to progress successfully with the remainder of the question. As usual in this type of question, there was an abundance of incorrect answers greater than 1 given.